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Chapter 8: Problem 19

A manufacturing process is designed to produce bolts with a 0.5-in. diameter. Once each day, a random sample of 36 bolts is selected and the diameters recorded. If the resulting sample mean is less than \(0.49\) in. or greater than \(0.51\) in., the process is shut down for adjustment. The standard deviation for diameter is \(0.02\) in. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an \(\bar{x}\) in the shutdown range when the true process mean really is \(0.5\) in.)

Short Answer

Expert verified
The probability of the manufacturing line being shut down unnecessarily is \(0.1336\) or \(13.36\% \).

Step by step solution

01

Define the Problem Variables

First, let's recognize the variables given in the problem - \(\mu = 0.5\), the true population mean. - \(\sigma = 0.02\), the true population standard deviation. - Sample size, \(n = 36\), the number of observations. We're looking for the probability that the sample mean \(\bar{x}\) falls outside the range \(0.49-0.51\).
02

Convert the Problem into a Standard Normal Distribution Problem

We're given a range where the production process is considered to be functioning correctly, \(0.49in < \bar{x} < 0.51in\). We must convert these to z-scores. To convert to z-scores, use the formula: \[Z = \frac{\bar{x} - \mu}{\frac{\sigma} {\sqrt{n}}}\] Where \(\bar{x} \) is the mean of our observations, \(\mu\) is the true mean, \(\sigma\) is the standard deviation, and \(n\) is the number of observations. In this case, transform the lower boundary (0.49in) and upper boundary (0.51in) to their corresponding z-values:
03

Compute the Lower Z-score:

Computation of lower z-score or z(lower boundary) provides: \[Z_{lower} = \frac{0.49 - 0.5}{\frac{0.02}{\sqrt{36}}} = -1.5\]
04

Compute the Upper Z-score:

Computation of upper z-score or z(upper boundary) provides: \[Z_{upper} = \frac{0.51 - 0.5}{\frac{0.02}{\sqrt{36}}} = 1.5\]
05

Calculate the Required Probability

We need to find the probability that the sample mean is outside this range, ie. \(P(Z < -1.5)\) or \(P(Z > 1.5)\). To find these, refer to the standard normal distribution table or calculator. The value corresponding to Z = 1.5 is \(0.4332\). Therefore, the probabilities we seek are : \[P(Z < -1.5) = P(Z > 1.5) = 0.5 - 0.4332\] Thus, the total probability of an unnecessary stoppage is given by two times the calculated probability.
06

Find the Result

Since we are looking for the manufacturing line to be shut down, either below the lower limit or above the upper limit. The probability of either event happening is calculated as: \[P = 2(0.5 - 0.4332) = 0.1336\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution that has a mean of 0 and a standard deviation of 1. It is a tool used in statistics to determine how data is dispersed around the mean, and it is fundamental to probability theory. When dealing with normal distributions that have different means and standard deviations, we can transform our measured values into z-scores to compare them to the standard normal distribution.

This conversion allows us to use standard normal distribution tables or calculators to find probabilities, which are essential for understanding and predicting outcomes in various fields, including manufacturing quality control. For example, the probability of a bolt鈥檚 diameter falling outside the acceptable range can be determined with the help of a standard normal distribution.
Sample Mean Calculation
The sample mean is an estimate of the population mean and is used to infer characteristics of the entire dataset from a smaller set of observations. To calculate the sample mean, you sum up all the observed values and then divide by the number of observations. In the context of manufacturing bolts, engineers take a sample to measure the mean diameter to decide whether the process needs adjustment. If the calculated sample mean diverges significantly from the expected process mean, it may indicate that something is amiss in the manufacturing process.

Accurate calculation of the sample mean is vital, as it allows quality control professionals to monitor process performance and take action when necessary to ensure that products meet specifications.
Z-score
A z-score, or standard score, is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. It is calculated by subtracting the mean from the value in question and then dividing the result by the standard deviation.

In a manufacturing context, determining the z-score for a sample mean helps to identify how far off the sample mean is from the process mean. If the z-score is within a certain range, the process may be considered in control. If not, the process might need adjustment, as indicated in our bolt diameter example. A z-score gives professionals a clear, standardized method for flagging potential issues and maintaining production quality.
Statistical Process Control
Statistical process control (SPC) is a methodological framework used for monitoring and controlling a process to ensure that it operates at its fullest potential. One of the key tools of SPC is the control chart, which displays data over time and helps detect any significant variation from the mean.

In the case of bolt manufacturing, SPC would involve regularly sampling the diameters of bolts and plotting the sample means on a control chart. An SPC analysis might reveal whether the process variation is due to common causes (random variation inherent to the process) or special causes (variation due to specific, identifiable factors). Properly applied, SPC can prevent unnecessary adjustments to the manufacturing process and help ensure consistent product quality.

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Most popular questions from this chapter

The article "Unmarried Couples More Likely to Be Interracial" (San Luis Obispo Tribune, March 13, 2002) reported that \(7 \%\) of married couples in the United States are mixed racially or ethnically. Consider the population consisting of all married couples in the United States. a. A random sample of \(n=100\) couples will be selected from this population and \(p\), the proportion of couples that are mixed racially or ethnically, will be computed. What are the mean and standard deviation of the sampling distribution of \(p\) ? b. Is it reasonable to assume that the sampling distribution of \(p\) is approximately normal for random samples of size \(n=100\) ? Explain. c. Suppose that the sample size is \(n=200\) rather than \(n=100\), as in Part (b). Does the change in sample size change the mean and standard deviation of the sampling distribution of \(p ?\) If so, what are the new values for the mean and standard deviation? If not, explain why not. d. Is it reasonable to assume that the sampling distribution of \(p\) is approximately normal for random samples of size \(n=200 ?\) Explain. e. When \(n=200\), what is the probability that the proportion of couples in the sample who are racially or ethnically mixed will be greater than \(.10\) ?

Newsweek (November 23, 1992) reported that 40\% of all U.S. employees participate in "self-insurance" health plans \((\pi=.40)\). a. In a random sample of 100 employees, what is the approximate probability that at least half of those in the sample participate in such a plan? b. Suppose you were told that at least 60 of the \(100 \mathrm{em}\) ployees in a sample from your state participated in such a plan. Would you think \(\pi=.40\) for your state? Explain.

Let \(x\) denote the time (in minutes) that it takes a fifth-grade student to read a certain passage. Suppose that the mean value and standard deviation of \(x\) are \(\mu=2 \mathrm{~min}\) and \(\sigma=0.8\) min, respectively. a. If \(\bar{x}\) is the sample average time for a random sample of \(n=9\) students, where is the \(\bar{x}\) distribution centered, and how much does it spread out about the center (as described by its standard deviation)? b. Repeat Part (a) for a sample of size of \(n=20\) and again for a sample of size \(n=100\). How do the centers and spreads of the three \(\bar{x}\) distributions compare to one another? Which sample size would be most likely to result in an \(\bar{x}\) value close to \(\mu\), and why?

Consider a population consisting of the following five values, which represent the number of video rentals during the academic year for each of five housemates: \(\begin{array}{lll} 8 & 14&16 & 10 & 11\end{array}\) a. Compute the mean of this population. b. Select a random sample of size 2 by writing the numbers on slips of paper, mixing them, and then selecting \(2 .\) Compute the mean of your sample. c. Repeatedly select samples of size 2 , and compute the \(\bar{x}\) value for each sample until you have the results of 25 samples. d. Construct a density histogram using the \(25 \bar{x}\) values. Are most of the \(\bar{x}\) values near the population mean? Do the \(\bar{x}\) values differ a lot from sample to sample, or do they tend to be similar?

The article "Thrillers" (Newsweek, April 22,1985 ) stated, "Surveys tell us that more than half of America's college graduates are avid readers of mystery novels." Let \(\pi\) denote the actual proportion of college graduates who are avid readers of mystery novels. Consider a sample proportion \(p\) that is based on a random sample of 225 college graduates. a. If \(\pi=.5\), what are the mean value and standard deviation of \(p ?\) Answer this question for \(\pi=.6\). Does \(p\) have approximately a normal distribution in both cases? Explain. b. Calculate \(P(p \geq .6)\) for both \(\pi=.5\) and \(\pi=.6\). c. Without doing any calculations, how do you think the probabilities in Part (b) would change if \(n\) were 400 rather than \(225 ?\)
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